[tex]1.)\\\\x^2 - 2x + 35 = 0\\\\a=1, \ \ b=-2, \ \ c=35 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{2-\sqrt{ (-2)^2-4 \cdot 1\cdot 35}}{2 }= \frac{2-\sqrt{4-140}}{2 }= \frac{2-\sqrt{-136}}{2 }=\\\\\frac{2-\sqrt{4*34}i}{2 }=\frac{2-2\sqrt{ 34}i}{2 }=\frac{2(1- \sqrt{ 34}i)}{2 }=1- \sqrt{ 34}i[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{2+\sqrt{ (-2)^2-4 \cdot 1\cdot 35}}{2 } =\frac{2-2\sqrt{ 34}i}{2 }=\frac{2(1+ \sqrt{ 34}i)}{2 }=1- \sqrt{ 34}i \\\\If\ \ b^{2}-4ac<0, \ then \ roots \ are \ imaginary \ (non-real)[/tex]
[tex]2.)\\\\x^2 + 2x =48\\\\x^2 + 2x -48=0\\\\a=1, \ \ b= 2, \ \ c=-48 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{-2-\sqrt{ 2^2-4 \cdot 1\cdot (-48)}}{2 }= \frac{-2-\sqrt{4+ 96}}{2 }= \frac{-2-\sqrt{100}}{2 }= \\\\=\frac{-2-10}{2 }=\frac{-12 }{2 }=-6[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-2+\sqrt{ 2^2-4 \cdot 1\cdot (-48)}}{2 }= =\frac{-2+10}{2 }=\frac{8 }{2 }=-4[/tex]
[tex]3.) \\\\4x + 32 = -x^2\\\\ x^2 +4x+32=0\\\\a=1, \ \ b= 4, \ \ c=32 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{-4-\sqrt{ 4^2-4 \cdot 1\cdot 32}}{2 }= \frac{-4-\sqrt{16-128}}{2 }= \frac{-4-\sqrt{-112}}{2 }= \\\\=\frac{-4- \sqrt{16*7}i}{2 }= \frac{-4- 4\sqrt{ 7}i}{2 }= \frac{2(-2- 2\sqrt{ 7}i)}{2 }= -2- 2\sqrt{ 7}i[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-4+\sqrt{ 4^2-4 \cdot 1\cdot 32}}{2 }= \frac{2(-2+ 2\sqrt{ 7}i)}{2 }= -2+ 2\sqrt{ 7}i \\\\If\ \ b^{2}-4ac<0, \ then \ roots \ are \ imaginary \ (non-real)[/tex]
